# 1、删除列表里所有字符串对象
# list1=[1, 2, 3, "str111","asdasd","qwdsa3d165",1,2,3,"asdqw","小天才","QA"]
# list2=[a for a in list1 if not isinstance(a,str)]
# print(list2)
# 2、删除列表中重复的元素
# list1=[1, 2, 3, "str111","asdasd","qwdsa3d165",1,2,3,"asdqw","小天才","QA"]
# l2=list(set(list1))
# print(l2)
# 3、l1 = [1, 2, 3, 4, 5]
# （1）在l1的元素3后面插入300
# l1 = [1, 2, 3, 4, 5]
# l1.insert(3,300)
# print(l1)
# （2）删除元素2
# l1 = [1, 2, 3, 4, 5]
# l1.remove(2)
# print(l1)
# （3）将5更改为500
# l1 = [1, 2, 3, 4, 5]
# l1[4]=500
# print(l1)
# （4）将2，3，4切片出来
# l1 = [1, 2, 3, 4, 5]
# print(l1[1:4])
# （5）l1[-3:-5]的结果
# l1 = [1, 2, 3, 4, 5]
# print(l1[-3:-5])
# （6）l1[-3:]的结果
# l1 = [1, 2, 3, 4, 5]
# print(l1[-3:])
#  2. 通过input引导用户输入一个姓名，判断该姓名是否存在于列表names中
# names = ["xi","eric","alvin","george"]
# name = input("name==")
# if name == ("xi","eric","alvin","george"):
#     print("yyy")
# else:
#     print(("nnn"))
# 4、l = [1,2,3,[4,5]]
# （1）将4修改为400
# l = [1,2,3,[4,5]]
# l[3]=[400,5]
# print(l)
# # （2）在l的[4，5]列表中追加一个6，即使l变为[1,2,3,[4,5,6]]
# l = [1,2,3,[4,5]]
# l[3] = [4,5,6]
# print(l)
# 5、'''小明去超市购买水果，账单如下
# 苹果  32.8
# 香蕉  22
# 葡萄  15.5
# 请将上面的数据存储到字典里，可以根据水果名称查询购买这个水果的费用
# 很简单哦，用水果名称做key，金额做value，创建一个字典
# dict1={}
# dict1.update({"苹果":"32.8","香蕉":"22","葡萄":"15.5"})
# print(dict1)
# 6、dic = {'python': 95,'java': 99,'c': 100}
# 字典的长度是多少
# dic = {'python': 95,'java': 99,'c': 100}
# print(len(dic))
# 请修改'java' 这个key对应的value值为98
# dic = {'python': 95,'java': 99,'c': 100}
# dic["java"]=98
# print(dic)
# 删除 c 这个key
# dic = {'python': 95,'java': 99,'c': 100}
# del dic["c"]
# print(dic)
# 增加一个key-value对，key值为 php, value是90
# dic = {'python': 95,'java': 99,'c': 100}
# dic["php"]=90
# print(dic)
# 获取所有的key值，存储在列表里
# dic = {'python': 95,'java': 99,'c': 100}
# print(dic.keys())
# 获取所有的value值，存储在列表里
# dic = {'python': 95,'java': 99,'c': 100}
# print(dic.values())
# 判断 javascript 是否在字典中
# dic = {'python': 95,'java': 99,'c': 100}
# print(dic["javascript"])
# 获得字典里所有value 的和
# dic = {'python': 95,'java': 99,'c': 100}
# print(sum(dic.values()))
# 获取字典里最大的value
# max1 = 0
# for i in dic1:
#     if max1 < i:
#         max = i
# print(max1)
# 获取字典里最小的value
# dic = {'python': 95,'java': 99,'c': 100}
# a = min(dic.values())
# print(a)
# 字典 dic1 = {'php': 97}， 将dic1的数据更新到dic中
# dic = {'python': 95,'java': 99,'c': 100}
# dic1 = {'php': 97}
# dic.update(dic1)
# print(dic)
# list1 = [12,45,78,12,45,78,12,77,99,53,6,93]
# 7、将重复的列表值加到一个新的字典里，key为数字，value为出现的次数
l1 = [12,45,78,12,45,78,12,77,99,53,6,93]
l2 = []
for i in l1:
    if l1.count(i)>1:
        l2.update({i:l1.count(i)})
print(l2)

# 8、求出列表里最大的数是多少（不允许用sort等其他内置函数去做）,以及这个数所在的索引位置
